Simplify the following expression: $\dfrac{49n}{35n^2}$ You can assume $n \neq 0$.
Explanation: $ \dfrac{49n}{35n^2} = \dfrac{49}{35} \cdot \dfrac{n}{n^2} $ To simplify $\frac{49}{35}$ , find the greatest common factor (GCD) of $49$ and $35$ $49 = 7 \cdot 7$ $35 = 5 \cdot 7$ $ \mbox{GCD}(49, 35) = 7 $ $ \dfrac{49}{35} \cdot \dfrac{n}{n^2} = \dfrac{7 \cdot 7}{7 \cdot 5} \cdot \dfrac{n}{n^2} $ $\phantom{ \dfrac{49}{35} \cdot \dfrac{1}{2}} = \dfrac{7}{5} \cdot \dfrac{n}{n^2} $ $ \dfrac{n}{n^2} = \dfrac{n}{n \cdot n} = \dfrac{1}{n} $ $ \dfrac{7}{5} \cdot \dfrac{1}{n} = \dfrac{7}{5n} $